How is a message $\mu$ encrypted in a witness encryption scheme below? What does C(x) = 1 mean below?

Paritosh007

3/10/24, 6:23 PM

I was reading this paper, where I came across the following statement on Pg 4 under the Homomorphic Commitments:

A witness encryption scheme, associated with a NP language, consists of an encryption and a decryption algorithm: Anyone can encrypt their message µ under an NP instance and the decryption algorithm can obtain µ using the witness to this instance. We use witness encryption as follows: The sender encrypts µ under the instance A · R_{C, x} + C(x) G which is obtained by homomorphically evaluating upon the commitments using the circuit C. The authority releases the decomitment R_C,x as witness which would then allow anyone to recover µ if and only if C(x) = 1.

where A is a uniformly sampled matrix from Z_q^{m x n}.

In the above statement, where is the message encryption done? Does it have anything to do with the matrix G in the statement ? What does C(x) = 1 actually mean in this statement? If its the output of the Circuit C on the committed value ( x ) which should be 1, why should that always be the case ?

Please Help.

0 + 1

multiparty-computation

homomorphic-encryption

garbled-circuits

Paritosh007

3/11/24, 5:35 PM

I think if C(x) = 0, then perhaps the trapdoor computation which is explained later in the paper won't be possible, and hence would prevent public decryption. Any add ons to this conclusion would be helpful.

Elon Musk

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EN: How is a message $\mu$ encrypted in a witness encryption scheme below? What does C(x) = 1 mean below?

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